mathwannabe
Junior Member
- Joined
- Feb 20, 2012
- Messages
- 122
Hello everybody 
I got this irrational equation I was working on... would be very happy if someone told me if I did it right.
1) How many solutions does the following equation have? \(\displaystyle \sqrt{3x+13}+\sqrt{x-1}=2\sqrt{x+3}\)
\(\displaystyle \sqrt{3x+13}+\sqrt{x-1}=2\sqrt{x+3}\)
\(\displaystyle (\sqrt{3x+13}+\sqrt{x-1})^2=(2\sqrt{x+3})^2\)
\(\displaystyle 3x+13+2\sqrt{(3x+13)(x-1)}+x-1=4(x+3)\)
\(\displaystyle 4x+12+2\sqrt{3x^2+10x-13}=4x+12\)
\(\displaystyle 2\sqrt{3x^2+10x-13}=0\)
\(\displaystyle (2\sqrt{3x^2+10x-13})^2=0\)
\(\displaystyle 4(3x^2+10x-13)=0\)
\(\displaystyle 3x^2+10x-13=0\)
\(\displaystyle x1=1\)
\(\displaystyle x2=-\dfrac{13}{3}\)
My answer is one... \(\displaystyle x1=1\)
I got this irrational equation I was working on... would be very happy if someone told me if I did it right.
1) How many solutions does the following equation have? \(\displaystyle \sqrt{3x+13}+\sqrt{x-1}=2\sqrt{x+3}\)
\(\displaystyle \sqrt{3x+13}+\sqrt{x-1}=2\sqrt{x+3}\)
\(\displaystyle (\sqrt{3x+13}+\sqrt{x-1})^2=(2\sqrt{x+3})^2\)
\(\displaystyle 3x+13+2\sqrt{(3x+13)(x-1)}+x-1=4(x+3)\)
\(\displaystyle 4x+12+2\sqrt{3x^2+10x-13}=4x+12\)
\(\displaystyle 2\sqrt{3x^2+10x-13}=0\)
\(\displaystyle (2\sqrt{3x^2+10x-13})^2=0\)
\(\displaystyle 4(3x^2+10x-13)=0\)
\(\displaystyle 3x^2+10x-13=0\)
\(\displaystyle x1=1\)
\(\displaystyle x2=-\dfrac{13}{3}\)
My answer is one... \(\displaystyle x1=1\)